LeetCode in Python

96. Unique Binary Search Trees

Medium

Given an integer n, return the number of structurally unique BST’s (binary search trees) which has exactly n nodes of unique values from 1 to n.

Example 1:

Input: n = 3

Output: 5

Example 2:

Input: n = 1

Output: 1

Constraints:

• 1 <= n <= 19

To solve this task using Python with a Solution class, you can follow these steps:

1. Define a class named Solution.
2. Inside the class, define a method named numTrees that takes n as an input parameter.
3. Implement an algorithm to calculate the number of structurally unique BSTs (binary search trees) with exactly n nodes.
4. Use dynamic programming to efficiently compute the number of unique BSTs.
5. Create a list dp of length n + 1 to store the number of unique BSTs for each number of nodes from 0 to n.
6. Initialize dp[0] and dp[1] to 1, as there is only one unique BST for empty tree or tree with only one node.
7. Iterate from 2 to n:
   • For each i, calculate dp[i] by summing up the products of dp[j - 1] and dp[i - j] for all possible values of j from 1 to i.
8. Return dp[n] as the result.

Here’s the implementation:

class Solution:
    def numTrees(self, n):
        dp = [0] * (n + 1)
        dp[0], dp[1] = 1, 1
        
        for i in range(2, n + 1):
            for j in range(1, i + 1):
                dp[i] += dp[j - 1] * dp[i - j]
        
        return dp[n]

# Example usage:
solution = Solution()
print(solution.numTrees(3))  # Output: 5
print(solution.numTrees(1))  # Output: 1

This solution efficiently calculates the number of structurally unique BSTs using dynamic programming. It iterates through each number of nodes from 2 to n and computes the number of unique BSTs for each number of nodes, resulting in a time complexity of O(n^2).

Solution

class Solution:
    def numTrees(self, n: int) -> int:
        def factorial(n):
            if n<1: return 1
            return n*factorial(n-1)
        fact = [factorial(i) for i in range(2*n+1)]
        return int(fact[2*n] / (fact[n] * fact[n] * (n+1)))

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